HDU 4405 Aeroplane chess(概率dp)
2016-07-13 16:54
337 查看
Aeroplane chess
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3119 Accepted Submission(s): 1993
[align=left]Problem Description[/align]
Hzz loves aeroplane chess very much. The chess map contains N+1 grids labeled from 0 to N. Hzz starts at grid 0. For each step he throws a dice(a dice have six faces with equal probability to face up and the numbers on the faces are
1,2,3,4,5,6). When Hzz is at grid i and the dice number is x, he will moves to grid i+x. Hzz finishes the game when i+x is equal to or greater than N.
There are also M flight lines on the chess map. The i-th flight line can help Hzz fly from grid Xi to Yi (0<Xi<Yi<=N) without throwing the dice. If there is another flight line from Yi, Hzz can take the flight line continuously. It is granted that there is
no two or more flight lines start from the same grid.
Please help Hzz calculate the expected dice throwing times to finish the game.
[align=left]Input[/align]
There are multiple test cases.
Each test case contains several lines.
The first line contains two integers N(1≤N≤100000) and M(0≤M≤1000).
Then M lines follow, each line contains two integers Xi,Yi(1≤Xi<Yi≤N).
The input end with N=0, M=0.
[align=left]Output[/align]
For each test case in the input, you should output a line indicating the expected dice throwing times. Output should be rounded to 4 digits after decimal point.
[align=left]Sample Input[/align]
2 0
8 3
2 4
4 5
7 8
0 0
[align=left]Sample Output[/align]
1.1667
2.3441
[align=left]Source[/align]
2012 ACM/ICPC Asia Regional Jinhua Online
[align=left]Recommend[/align]
zhoujiaqi2010
题目大意:
玩飞行棋,有一个骰子可以走1~6步,还有一些航线可以直接从一个格子移动到前面的一个格子。求走到终点的步数期望。
解题思路:
从终点到起点递推,如果一个格子有航线,那么它的期望就是航线所连接的前面的格子的期望,否则它的期望就是前面6步格子期望除以6再加一。
附AC代码:
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=100000+5;
int path[maxn],n,m;
double dp[maxn];
int main()
{
while(~scanf("%d%d",&n,&m)&&(n||m))
{
memset(path,-1,sizeof path);//-1表示没有航线
for(int i=0;i<m;++i)
{
int from;
scanf("%d",&from);
scanf("%d",&path[from]);
}
memset(dp,0,sizeof dp);//初始期望为0
for(int i=n-1;i>=0;--i)
{
if(path[i]!=-1)//有航线
dp[i]=dp[path[i]];
else
{
for(int j=1;j<=6;++j)
dp[i]+=dp[i+j];
dp[i]=dp[i]/6+1;
}
}
printf("%.4f\n",dp[0]);
}
return 0;
}
相关文章推荐
- 高并发电子商务平台技术架构
- 技术那点事-中文IT信息服务网站,为IT专业技术人员提供最全面的信息和服务的网站
- Linux文章汇总
- 多个tomcat之间的session复制
- 大型网站架构
- 使用OPENGL绘制一个带轨迹的小球
- 【终端快捷键】Linux terminal 终端常用快捷键
- PopUpWindow使用详解(一)——基本使用
- drop、delete和truncate三者的区别
- Suse linux 11 SP3安装VMWare Tools时问题的解决
- Linux基础(4)之文本处理
- 刚开始添加Open CV的时候遇到问题
- Openjudge-NOI题库-和为给定数
- linux系统编程手册 定时器和休眠
- Hadoop的发行版本介绍
- org.apache.struts.action.InvalidCancelException异常解决方法
- Linux grep使用详解
- Jcrop使用方法
- nginx 编译安装
- Linux 常用命令